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Decay Constant & Half-Life (HL) (HL IB Physics)

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Katie M

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Katie M

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Physics

Decay Constant & Half-Life

  • Since radioactive decay is spontaneous and random, it is useful to consider the average number of nuclei that are expected to decay per unit time
    • This is known as the average decay rate

  • As a result, each radioactive element can be assigned a decay constant
  • The decay constant λ is defined as:

 The probability that an individual nucleus will decay per unit of time

  • When a sample is highly radioactive, this means the number of decays per unit time is very high
    • This suggests it has a high level of activity

  • Activity, or the number of decays per unit time can be calculated using:

  • Where:
    • A = activity of the sample (Bq)
    • ΔN = number of decayed nuclei
    • Δt = time interval (s)
    • λ = decay constant (s-1)
    • N = number of nuclei remaining in a sample

  • The activity of a sample is measured in Becquerels (Bq)
    • An activity of 1 Bq is equal to one decay per second, or 1 s-1

  • This equation shows:
    • The greater the decay constant, the greater the activity of the sample
    • The activity depends on the number of undecayed nuclei remaining in the sample
    • The minus sign indicates that the number of nuclei remaining decreases with time

  • Half-life and decay constant can be linked, using an equation called the exponential decay equation showing how the number of active nuclei, N, changes over time, t, where N0 is the initial number of active nuclei

N = N0 e–λt

  • When time t is equal to the half-life t½, the number of active nuclei in the sample, N, will be half of its original value, so N = ½ N0

Calculating Half-Life equation 1

  • The formula linking half-life and decay constant can then be derived as follows:

Calculating Half-Life equation 2

Calculating Half-Life equation 3

Calculating Half-Life equation 3a

  • Therefore, half-life t½ can be calculated using the equation:

Calculating Half-Life equation 4

  • This equation shows that half-life t½ and the radioactive decay rate constant λ are inversely proportional
    • Therefore, the shorter the half-life, the larger the decay constant and the faster the decay
  • The half-life of a radioactive substance can be determined from decay curves and log graphs
  • Since half-life is the time taken for the initial number of nuclei, or activity, to reduce by half, it can be found by
    • Drawing a line to the curve at the point where the activity has dropped to half of its original value
    • Drawing a line from the curve to the time axis, this is the half-life

Half Life Decay Curves 1, downloadable AS & A Level Physics revision notes

A linear decay curve. This represents the relationship: fraction numerator increment N over denominator increment t end fraction equals negative lambda N

  • Straight-line graphs tend to be more useful than curves for interpreting data
    • Due to the exponential nature of radioactive decay logarithms can be used to achieve a straight line graph

  • Take the exponential decay equation for the number of nuclei

N = N0 e–λt

  • Taking the natural logs of both sides

ln N = ln (N0) − λt

ln N = − λt + ln (N0

  • In this form, this equation can be compared to the equation of a straight line

y = mx + c

  • Where:
    • ln (N) is plotted on the y-axis
    • t is plotted on the x-axis
    • gradient = −λ
    • y-intercept = ln (N0)
  • Half-lives can be found in a similar way to the decay curve but the intervals will be regular as shown below:

Half Life Decay Curves 2, downloadable AS & A Level Physics revision notes

A logarithmic graph. This represents the relationship: ln space N equals negative lambda t plus ln space N subscript 0

  • Note: experimentally, the measurement generally taken is the count rate of the source
    • Since count rate ∝ activity ∝ number of nuclei, the graphs will all take the same shapes when plotted against time (or number of half-lives) linearly or logarithmically  

Worked example

Radium is a radioactive element first discovered by Marie and Pierre Curie.

They used the radiation emitted from radium-226 to define a unit called the Curie (Ci) which they defined as the activity of 1 gram of radium.

It was found that in a 1 g sample of radium, 2.22 × 1012 atoms decayed in 1 minute.Another sample containing 3.2 × 1022 radium-226 atoms had an activity of 12 Ci.

(a) Determine the value of 1 Curie

(b) Determine the decay constant for radium-226

Answer:

(a)

Step 1: Write down the known quantities

  • Number of atoms decayed, ΔN = 2.22 × 1012
  • Time, Δt = 1 minutes = 60 s

Step 2: Write down the activity equation

Step 3: Calculate the value of 1 Ci

(b)

Step 1: Write down the known quantities

  • Number of atoms, N = 3.2 × 1022
  • Activity, A = 12 Ci = 12 × (3.7 × 1010) = 4.44 × 1011 Bq

Step 2: Write down the activity equation

A = λN

Step 3: Calculate the decay constant of radium

  • Therefore, the decay constant of radium-226 is 1.4 × 10–11 s–1

Worked example

Strontium-90 is a radioactive isotope with a half-life of 28.0 years. A sample of Strontium-90 has an activity of 6.4 × 109 Bq.

(a)
Calculate the decay constant λ, in year–1, of Strontium-90.
(b)
Determine the fraction of the sample remaining after 50 years.

 

Answer:

(a)

Step 1: List the known quantities

  • Half-life, t½ = 28 years

Step 2: Write the equation for half-life

t subscript bevelled 1 half end subscript equals fraction numerator ln space 2 over denominator lambda end fraction

Step 3: Rearrange for λ and calculate

lambda equals fraction numerator ln space 2 over denominator t subscript bevelled 1 half end subscript end fraction equals fraction numerator ln space 2 over denominator 28 end fraction= 0.025 year−1

(b)

Step 1: List the known quantities

  • Decay constant, λ = 0.025 year−1
  • Time passed, t = 50 years

Step 2: Write the equation for exponential decay

N equals N subscript 0 space e to the power of negative lambda t end exponent

Step 3: Rearrange for N over N subscript 0 and calculate

N over N subscript 0 equals e to the power of negative lambda t end exponent

N over N subscript 0 equals e to the power of negative left parenthesis 0.025 right parenthesis cross times 50 end exponent = 0.287

  • Therefore, 28.7% of the sample will remain after 50 years

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Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.